Caluclate the delta H for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is -146.8kJ/mol. Balanced equation balanced equation:
C5H12(l)+8O2(g)-->5CO2+6H2O(g)
CO2=393.5
H2O=40.7
Well, to calculate the standard enthalpy change (ΔH) for this reaction using the standard enthalpies of formation, we need to subtract the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products.
So, let's calculate it step by step:
Reactants:
C5H12(l): -146.8 kJ/mol (given)
O2(g): 0 kJ/mol (oxygen gas is in its standard state)
Products:
5CO2(g): (5)(393.5 kJ/mol) = 1967.5 kJ/mol
6H2O(g): (6)(40.7 kJ/mol) = 244.2 kJ/mol
Now, let's sum up the products and reactants:
Sum of reactants: -146.8 kJ/mol + 0 kJ/mol = -146.8 kJ/mol
Sum of products: 1967.5 kJ/mol + 244.2 kJ/mol = 2211.7 kJ/mol
Finally, calculate ΔH:
ΔH = sum of products - sum of reactants
ΔH = 2211.7 kJ/mol - (-146.8 kJ/mol)
ΔH = 2211.7 kJ/mol + 146.8 kJ/mol
ΔH ≈ 2358.5 kJ/mol
So, the ΔH for this reaction is approximately 2358.5 kJ/mol.
That's a lot of heat! I hope the reaction doesn't get too hot, or else things might start melting around here. Stay cool!
To calculate the ΔH for the reaction using standard enthalpies of formation, we need to use the balanced equation and the standard enthalpies of formation for each compound involved.
Given:
Standard enthalpy of formation of liquid pentane (C5H12): -146.8 kJ/mol
Standard enthalpy of formation of CO2: 393.5 kJ/mol
Standard enthalpy of formation of H2O: 40.7 kJ/mol
The balanced equation is:
C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)
Step 1: Calculate the ΔH for the products.
ΔH for CO2 = (5 mol) * (393.5 kJ/mol) = 1967.5 kJ
ΔH for H2O = (6 mol) * (40.7 kJ/mol) = 244.2 kJ
Step 2: Calculate the ΔH for the reactants.
ΔH for C5H12 = (1 mol) * (-146.8 kJ/mol) = -146.8 kJ
Step 3: Calculate the ΔH for the reaction using the equation:
ΔH = ΣΔH(products) - ΣΔH(reactants)
ΔH = (ΔH for CO2 + ΔH for H2O) - (ΔH for C5H12)
ΔH = (1967.5 kJ + 244.2 kJ) - (-146.8 kJ)
ΔH = 2211.7 kJ + 146.8 kJ
ΔH = 2358.5 kJ
Therefore, the ΔH for the reaction is 2358.5 kJ.
To calculate the ΔH for the given reaction using standard enthalpies of formation, you need to follow these steps:
Step 1: Write down the balanced equation for the reaction:
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)
Step 2: Calculate the ΔH for the reaction by using the standard enthalpies of formation for each compound involved. The standard enthalpy of formation (∆Hf°) is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states at 298 K and 1 atm pressure.
Given the standard enthalpy of formation values:
∆Hf° (C5H12) = -146.8 kJ/mol
∆Hf° (CO2) = 393.5 kJ/mol
∆Hf° (H2O) = 40.7 kJ/mol
Step 3: Determine the net ΔH by taking into account the stoichiometric coefficients of the balanced equation. The ΔH for a reaction is the sum of the products' enthalpies minus the sum of the reactants' enthalpies.
ΔHrxn = (5 × ∆Hf° (CO2)) + (6 × ∆Hf° (H2O)) - (∆Hf° (C5H12) + 8 × ∆Hf° (O2))
Step 4: Substitute the known values into the equation and calculate:
ΔHrxn = (5 × 393.5) + (6 × 40.7) - (-146.8 + 8 × 0)
ΔHrxn = 1967.5 + 244.2 + 146.8
ΔHrxn ≈ 2358.5 kJ
Therefore, the ΔH for the given reaction using standard enthalpies of formation is approximately 2358.5 kJ.